The bass in clear Lake have weights that are normally distributed with a mean of
ID: 3152246 • Letter: T
Question
The bass in clear Lake have weights that are normally distributed with a mean of 2.2 lbs and a stardard deviation of 0.7 lbs.
(a) if you catch one random bass from Clear lake, find the probability that it weighs less than 1.25 pounds?
(b)if you go out and catch six bass one day from Clear Lake. find the probability that their mean weight is less than
1.25 pounds
(c) if you catch one random bass from Clear lake, find the probability that it weighs more than 4 pounds
(d) suppose you only want to keep
fish that are in the top 80% as far as weight is concerned. what is the minimum weight of a bass to be a keeper?
Explanation / Answer
MEAN = 2.2
STANDARD DEVIATION = 0.7
THE DISTRIBUTION IS NORMAL THEREFORE WE CAN USE THE FORMULA
Z = (X-MEAN)/STANDARD DEVIATION
A) HERE N = 1
P(X<1.25) =
For x = 1.25, the z-value z = (1.25 - 2.2) / 0.7 = - 1.35
Hence P(x < 1.25) = P(z < -1.35), now from the z table we will take the value of z score = -1.35
And that value will be the probability required.
= [area to the left of -1.35] = 0.0885
B) HERE N = 6
THEREFORE WE WILL USE THE FORMULA
Z = (X-MEAN)/(STANDARD DEVIATION/SQRT(N)
P(X<1.25) =
For x = 1.25, the z-value z = (1.25 - 2.2) / (0.7/SQRT(6)) = -3.3
Hence P(x < 1.25) = P(z < -3.3), now from the z table we will take the value of z score = -3.3
And that value will be the probability required.
= [area to the left of -3.3] = 0.0005
C) P(X>4)=
For x = 4, z = (4 - 2.2) / 0.7 = 2.57
Hence P(x > 4) = P(z > 2.57) = [total area] - [area to the left of 2.57]
1 - [area to the left of 2.57]
now from the z table we will take the value of z score = 2.57
= 1 - 0.9949 = 0.0051
D) THE FORMULA TO BE USED
X = MEAN+Z*STANDARD DEVIATION
Z SCORE FOR 80% = 1.28
THEREFORE
X = 2.2+1.28*0.7
THEREFORE X = 3.096
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